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Imagine, if you will, there exists somewhere out there a supermassive planet. For some reason, this planet is only a shell of its former existence and all that is left is a crust of substantial size to keep it from collapsing in on itself (let's imagine it is stable at 5000' thick). Now, this planet was HUGE! How large, you ask? Large enough that the remaining shell is still able to maintain a gravity well strong enough to keep you anchored to the surface.

Pretty neat, huh?

Well, since this is a sight to be seen by everyone, you stroll along and, not looking where you are going, you tumble into a chasm and straight down into a hole that permeates the thickness of the shell. As you are falling down and down and down, you think:

"At what point do I stop falling down and I start falling up? And would I be able to retain the velocity, since gravity is a conservative force, and fall back through another hole on the surface on the opposite side? Where would the center of gravity even exist? I mean, aaaaahhhhhhh!"

What say you, browsers and scientists? A stretch of the imagination, sure, but is it physically possible? Could there exists a configuration similar to this where gravity could be concentrated to a point but still allow free travel up to the focus?

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    $\begingroup$ Emilio Pisanty's answer is a great one to this particular answer. But, from a materials perspective, it would be pretty much impossible for a planet like this to form unless very small. The stablest (lowest energy) configuration is a spherical solid, which is why celestial bodies over a certain size are spherical. Over time hollows would slump in on themselves (stress, asteroid impact). A structure like yours would be a built structure lasting only civilization-length timescales. Anyone for a coffee at the Death Star Canteen? Great question BTW $\endgroup$ – WetSavannaAnimal May 8 '15 at 0:15
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    $\begingroup$ @WetSavannaAnimalakaRodVance: Concerning the just-deleted answer, it doesn't work even for a solid planet (with a single vertical tunnel to fall in). You need a non-conservative force to change the total (kinetic+potential) energy, all gravity ever does is convert between the two. $\endgroup$ – Ben Voigt May 8 '15 at 0:55
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    $\begingroup$ @Rincewind Nothing to do with the solution here, but this question asked under that username makes me happy. $\endgroup$ – Asher May 8 '15 at 1:03
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    $\begingroup$ If you had a planet in such a configuration, note that the atmosphere would get sucked into the center (through the hole you fell in, if nothing else). So inside the shell you would experience (slight) gravitational pull toward the center, and potentially a lot of air resistance, depending on how much atmosphere the planet has. $\endgroup$ – Aaron Dufour May 8 '15 at 14:31
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Yes, this is possible. It is perfectly fine for a mass configuration to

  • produce, for points outside a sphere of radius $R$ centred at $\mathbf r_0$, a gravitational field identical to that of a point mass at $\mathbf r_0$, and still

  • be completely empty inside a smaller sphere of radius $a$ around $\mathbf r_0$.

The spherical-shell model you describe is the simplest example of this. (On the other hand, it does not describe a physically realizable model from a materials standpoint, as described in Rod Vance's answer.)

In addition to this, your spherical shell of a planet will have the additional property that once you are inside the shell, the planet's gravitational attraction will vanish. Objects there will perform uniform linear motion with constant velocity, until of course they reach the edge of the shell.

This has been known since the time of Newton and it is one of the standard exercises in electrostatics (which is mathematically identical to newtonian gravity) for a physics undergrad. Schemes like this are fairly common and relatively fun to analyse; you may find the "gravity train" interesting.

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    $\begingroup$ Wow! That is amazing! So, there is an official line of demarkation for gravity that exists at the crust. I wonder why the well extends outward but not inward and I wonder how this could be exploited to benefit us. $\endgroup$ – Rincewind May 7 '15 at 19:51
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    $\begingroup$ For a pure shell, yes, the gravitational well does not get deeper inside the shell (it is still deep, it's just flat). However, we don't have a lot of those shells around so the fact is of little practical use. Also note that gravity goes down linearly to zero, so it doesn't suddenly disappear - it just gradually tails off. $\endgroup$ – Emilio Pisanty May 7 '15 at 19:58
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    $\begingroup$ If you are inside the sphere, but not at the centre, why don't you feel some attraction towards the centre? For example - if you're halfway between the centre and the shell, there would be more shell (and therefore more mass) in one direction than the other - so why doesn't it create a difference in gravitational attraction? $\endgroup$ – HorusKol May 8 '15 at 3:58
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    $\begingroup$ @HorusKol The thing is that for each element of solid angle, on one side there is more shell (and therefore more mass), but on the other side the shell is closer (and therefore the attraction is stronger). Both effects scale as $1/r^2$ and they exactly cancel out. See any introductory electrostatics textbook for the details. $\endgroup$ – Emilio Pisanty May 8 '15 at 7:54
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    $\begingroup$ @Rincewind You lost me, I'm afraid. What do you mean by focus? The gravitational field for a shell goes look different on either side of the shell, as I've explained. $\endgroup$ – Emilio Pisanty May 8 '15 at 7:59
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Emilio Pisanty's answer is a great one to this particular answer, i.e. there is no in principle bar from the laws of physics to a structure like yours and it would have the zero gravity inside property that Emilio describes.

But, from a materials perspective, it would be pretty much impossible for a planet like this to form unless very small. The stablest (lowest energy) configuration of a clump of matter is a spherical solid, which is why celestial bodies over a certain size are spherical. Over time hollows would slump in on themselves owing to stress, asteroid impact and the like. Smaller bodies can be nonspherical owing their strength (characterized by quantities such as tensile strength) i.e. the internal forces that they can support before being shorn apart and "flow" like a fluid are nonzero. They will "flow" until all internal forces are compressive: materials tend to be much stronger in compression than tension. But this limit is reached for the kinds of matter that asteroids and planets are made of pretty quickly. Take a look at this table of Solar System moons from NASA and check their dimensions: this will give you a good idea of this limit. Anything with dimensions of several hundred kilometers or more is listed as spherical. A good example is Ceres: a highly spherical asteroid (and the biggest asteroid). Below is a scale comparison of the asteroids Ceres, Vesta and Eros and it gives a good, qualitative feel for the roundness versus size relationship:

Ceres, Vesta and Eros

Source Wikipedia page "Ceres (Dwarf Planet)"

These are solid structures. The maximum spherical size for a hollow structure would be less.

A structure like yours would be a built structure lasting only civilization-lifetime-length timescales: probably much less than one million years.

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    $\begingroup$ So what is that bright spot in the middle of Ceres, then? $\endgroup$ – Emilio Pisanty May 8 '15 at 8:56
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    $\begingroup$ @EmilioPisanty According to Wikipedia: "A bright spot seen in earlier Hubble images was resolved as two distinct high-albedo features inside a crater in a 19 February 2015 image, leading to speculation about a possible cryovolcanic origin[24][25][26] or outgassing.[27] On 3 March 2015, a NASA spokesperson said the spots are consistent with highly reflective materials containing ice or salts, but that cryovolcanism is unlikely.[28]" $\endgroup$ – Random832 May 8 '15 at 13:15
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According to the shell theorem, the gravitational force inside such a hollow spherical planet would be 0

On the outside, gravitational force would be as if the planet was a regular planet. At some point during your fall, the gravitational force pulling you to the center would decrease to 0. If there's atmosphere inside the planet, it would slow you down until you came to a stop eventually. Without an atmosphere to stop you, according to the theorem there's no force to change your movement and you will fall to the other side of the shell.

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  • $\begingroup$ What would happen if the diameter of the planet were reduced while the thickness of the crust is increased until the gravitational fields permeating on the inside overlapped and created a focal point? Wouldn't that emulate (to a degree) what a solid body of mass is doing? I imagine it would change oscillations if one were to fall through the planet. $\endgroup$ – Rincewind May 7 '15 at 20:38
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    $\begingroup$ @Rincewind what do you mean by "focal point" ? $\endgroup$ – Name May 7 '15 at 20:44
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    $\begingroup$ @Rincewind, Re, until the gravitational fields permeating on the inside overlapped... Doesn't make any sense. Inside your hollow planet, no matter how thick or how thin the shell, you experience a gravitational pull from every atom in the shell. They all overlap, but they all pull you in different directions. It can be mathematically shown that the net pull at any point within the hollow space is zero. $\endgroup$ – Solomon Slow May 7 '15 at 21:34
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    $\begingroup$ @Ricewind There is no "limit of influence." The magnitude $f$ of the gravitaional attraction beteween two masses $m_1$ and $m_2$ separated by distance $r$ is $\frac{Gm_1m_2}{r^2}$ (Newton's Universal Law of Gravitation). Search for "shell theorem" in Wikipedia for a mathematical explanation of how that law manifests itself inside your hollow planet. $\endgroup$ – Solomon Slow May 8 '15 at 16:53
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    $\begingroup$ @Rincewind noooo, you will probably die! Oh, wait... you mean to search wikipedia...not the jumping into a hollow planet business. $\endgroup$ – Name May 11 '15 at 20:38
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What if this planet existed and you fell in? Let's take a look at that and see.

I'll spare listing all the reasons why a planet like this can't exist and wouldn't last long if it did; those are no fun. So instead, let's say this planet does, for some reason, exist in some galaxy far far away and it's made of something strong enough that it'll be around for the long haul (So that's where all the adamantium went!). As stated elsewhere, the shell theorem says that the gravity inside this big, metal, hollow sphere should drop to zero. This happens because the distribution of matter on all sides has a net canceling effect on the gravitational force felt by anything inside the shell. But now we're assuming there's absolutely nothing inside the planet shell. Is this really realistic? (that is, is it realistic for our unrealistic plant-shell scenario? Because if it must be unrealistic, it should be at least realistically unrealistic. I think that's a realistic requirement).

So here you are, strolling around on this planet shell because, as you say, that's a neato-burrito find (that's the technical term, I believe). Is there air? Or are you in a spacesuit? If there's air, we have a problem. You said this shell has a hole you inevitably fall through. How big is this hole? 2 meters across? (Why, that's no bigger than a womp rat). That means the atmosphere is linked to the vacuum inside the shell. I'll give you two guesses what happens next; it isn't pretty. The pressure from the atmosphere would rapidly push air into the interior of the shell using the hole as a massive Venturi tube. Not to mention the decreasing gravity on the way down would continue to increase the speed of the air molecules and, thus, pull air from the atmosphere faster. I'm no doctor, but I can tell you that would make a pretty big shockwave through the atmosphere. You probably wouldn't live through it (pfft! Not with THAT attitude!). If you did, what you'd find after an equilibrium is reached is that now the shell isn't empty. As such, not only would you have air resistance to deal with when you fall through the hole, but there would be a net gravitational effect inside the shell due to the mass of air it contains (albeit, not nearly as strong as the gravity outside the shell). The result? You aren't coming back out. You'll oscillate a bit and come to rest at the center of the sphere. Or you'll be annihilated when the atmosphere rushed into the shell, but let's not dwell on unpleasant things.

Okay, you say, what if I were in a spacesuit and there was no atmosphere? Phew. I've got just one problem now. You're now in a galaxy far far away, on a giant metal sphere about the size of a small moon that's mostly hollow inside and has a 2 metre shaft that leads directly to its core. You should expect regular gravity everywhere inside because, contrary to what the shell theorem says in this case, I'm fairly certain the Death Star had artificial gravity generators.

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  • $\begingroup$ Haha! Fantastic answer! Answered in almost the same fashion in which it was asked. $\endgroup$ – Rincewind May 9 '15 at 1:00
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Take a baseball and put it at about ground level and drop it into the hole.

The gravity will pull on the ball and will accelerate and plunge through the planet, the void, and on it's way up again (or down from your perspective), gravity will slow it's ascent to the point that it'll just about make it to ground level on the other. The ball will spring back up(?) to you but it'll just make it out of your reach. If you let it fall back down(?) (here it comes again, and again, and again) it should eventually settle in the center of the hollow planet.

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    $\begingroup$ What you describe would be true for a ball dropped down a narrow borehole through the center of a solid planet, but if there was a spherical void concentric with the center of the planet, the ball would experience no acceleration anywhere within the void. $\endgroup$ – Solomon Slow May 8 '15 at 16:53
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    $\begingroup$ As I understand, there gravitational attraction is centered on the matter (the crust) rather than the center of the planet. Once inside, the forces cancel each other out and there is no attraction experienced at all. $\endgroup$ – Rincewind May 8 '15 at 18:43

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